
How do you find the exact value of \[\arcsin \left( \dfrac{1}{3} \right)\] ?
Answer
544.8k+ views
Hint: Here arcsine is used to denote the inverse of sine which is represented as ${{\sin }^{-1}}$ . The sine of any angle is the ratio of length of opposite side to the length of hypotenuse in a right-angle triangle. For example, graphically $\sin \left( \dfrac{\pi }{2} \right)=1$ . Since here we have to find arcsin of a transcendental value therefore to find the exact value, we need a calculator.
Complete answer:
In the question we have to find arcsin. Arcsin or sine inverse is sine inverse of a sine expression which gives the value of angle. It is equal to the ratio of length of hypotenuse to the length of the opposite side in a right-angle triangle. For example, ${{\sin }^{-1}}(1)=\dfrac{\pi }{2}$ .
We know that $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ so sine inverse of half will be equal to, ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6}$ . Here we are asked to find \[\arcsin \left( \dfrac{1}{3} \right)\] which is ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)$ . So, we can say that the value of ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)$ must be less than $\dfrac{\pi }{6}$ . But we know only standard radian values of trigonometric functions which are multiples of $0,\dfrac{\pi }{6},\dfrac{\pi }{3},\dfrac{\pi }{2}$ . But here we are asked to find a value which will be less than $\dfrac{\pi }{6}$ but greater than $0$ so the value will be in the first quadrant. So, we will have to use a calculator to find the exact value of ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)$ . Enter the inverse sine mode in the calculator, enter the value of the number then obtain the result in radian or degree mode.
So, the value of ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)$ comes out to be $0.339836$ rad.
Note: While solving the question of this type for non-transcendental values keep in mind the range and domain of the trigonometric functions. In the above question we can’t find the value by direct calculations therefore, we will use a calculator to find the correct and exact value. Also, when we are finding values of trigonometric functions, it is better to use a graph of that function.
Complete answer:
In the question we have to find arcsin. Arcsin or sine inverse is sine inverse of a sine expression which gives the value of angle. It is equal to the ratio of length of hypotenuse to the length of the opposite side in a right-angle triangle. For example, ${{\sin }^{-1}}(1)=\dfrac{\pi }{2}$ .
We know that $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ so sine inverse of half will be equal to, ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6}$ . Here we are asked to find \[\arcsin \left( \dfrac{1}{3} \right)\] which is ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)$ . So, we can say that the value of ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)$ must be less than $\dfrac{\pi }{6}$ . But we know only standard radian values of trigonometric functions which are multiples of $0,\dfrac{\pi }{6},\dfrac{\pi }{3},\dfrac{\pi }{2}$ . But here we are asked to find a value which will be less than $\dfrac{\pi }{6}$ but greater than $0$ so the value will be in the first quadrant. So, we will have to use a calculator to find the exact value of ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)$ . Enter the inverse sine mode in the calculator, enter the value of the number then obtain the result in radian or degree mode.
So, the value of ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)$ comes out to be $0.339836$ rad.
Note: While solving the question of this type for non-transcendental values keep in mind the range and domain of the trigonometric functions. In the above question we can’t find the value by direct calculations therefore, we will use a calculator to find the correct and exact value. Also, when we are finding values of trigonometric functions, it is better to use a graph of that function.
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